If \[ \sec(x+\alpha)+\sec(x-\alpha)=2\sec x, \] prove that \[ \cos x=\pm \sqrt{2}\cos\frac{\alpha}{2} \]
Question
If \[ \sec(x+\alpha)+\sec(x-\alpha)=2\sec x, \] prove that \[ \cos x=\pm \sqrt{2}\cos\frac{\alpha}{2}. \]
Solution
Given,
\[ \sec(x+\alpha)+\sec(x-\alpha)=2\sec x \]
Converting secant into cosine,
\[ \frac{1}{\cos(x+\alpha)} + \frac{1}{\cos(x-\alpha)} = \frac{2}{\cos x} \]
Taking LCM on the left side,
\[ \frac{ \cos(x-\alpha)+\cos(x+\alpha) }{ \cos(x+\alpha)\cos(x-\alpha) } = \frac{2}{\cos x} \]
Using the identity
\[ \cos C+\cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2} \]
we get
\[ \cos(x-\alpha)+\cos(x+\alpha) = 2\cos x\cos\alpha \]
Therefore,
\[ \frac{ 2\cos x\cos\alpha }{ \cos(x+\alpha)\cos(x-\alpha) } = \frac{2}{\cos x} \]
Cross multiplying,
\[ 2\cos^2x\cos\alpha = 2\cos(x+\alpha)\cos(x-\alpha) \]
Cancelling 2,
\[ \cos^2x\cos\alpha = \cos(x+\alpha)\cos(x-\alpha) \]
Using the identity
\[ \cos(A+B)\cos(A-B) = \cos^2A-\sin^2B \]
we get
\[ \cos^2x\cos\alpha = \cos^2x-\sin^2\alpha \]
Rearranging,
\[ \cos^2x-\cos^2x\cos\alpha = \sin^2\alpha \]
\[ \cos^2x(1-\cos\alpha) = \sin^2\alpha \]
Using
\[ 1-\cos\alpha = 2\sin^2\frac{\alpha}{2} \]
and
\[ \sin\alpha = 2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2} \]
Therefore,
\[ \cos^2x \cdot 2\sin^2\frac{\alpha}{2} = 4\sin^2\frac{\alpha}{2}\cos^2\frac{\alpha}{2} \]
Cancelling \[ 2\sin^2\frac{\alpha}{2} \] from both sides,
\[ \cos^2x = 2\cos^2\frac{\alpha}{2} \]
Taking square root,
\[ \boxed{ \cos x = \pm \sqrt{2}\cos\frac{\alpha}{2} } \]
Hence proved.